Working out Riemann sums

Question

Consider $f:\mathbb{R} \to \mathbb{R}$, $f(x) = x$ on an interval $[-a, a]$.

Then my Riemann sum becomes:

$$\displaystyle S = \sum^{n-1}_{k = 0}a_{k}[x_{k} - x_{k - 1}]$$

where $a_{k}$ is the internal point of the subinterval $[x_{k-1}, x_{k}]$ of the partition $P = \{ x_{0}, x_{1}, ..., x_{n} \}$. How do I progress with this?

I'm not sure how to do a Riemann Integral of a rational function... — DonAntonio, Feb 05 '14 at 18:49
@DonAntonio, what about the real function $f(x) = x$ for rational $x$ and $f(x) = 0$ for irrational $x$, I was able to do the latter so just wrote up the first. — user2850514, Feb 05 '14 at 19:03
That is fine, @user2850514 : the example you give is a function defined for all real numbers. Your question's data is a function from the rationals to the rationals...very different. — DonAntonio, Feb 05 '14 at 19:38
@user2850514 I've no idea. Why is that function of yours defined only on the rationals? And if it for whatever reason, why would you want to do Riemann Summs with it? — DonAntonio, Feb 05 '14 at 19:51
@DonAntonio Well the function as a whole is $f(x) = x$ for rational $x$ and $f(x) = 0$ for irrational $x$, I want to show that the Riemann integral of an arbitrary interval $[-\delta, \delta]$ does not exist, to do so I am working them out separately and showing a contradiction. — user2850514, Feb 05 '14 at 19:54
Oooh, now I see ! Why didn't you write so in the question?! — DonAntonio, Feb 05 '14 at 19:55

DonAntonio · Accepted Answer · 2014-02-06T10:21:43.687

0

For any subinterval of any partition of the interval $\;[-\delta,\delta]\;$, we can always choose one $\;0<a_l\in\Bbb Q\;$ and all the rest of the $\;a_k\in\Bbb R\setminus\Bbb Q\;$ , so in this case we get $$\sum_{k=1}^n a_k(x_k-x_{k-1})=a_l(x_l-x_{l-1})>0$$

On the other hand, if all the $\;a_i\in\Bbb R\setminus\Bbb Q\;$ , then the sum is obviously zero. This proves the Riemann sums do not approach any definite limit and thus the function isn't integrable.

edited Feb 06 '14 at 10:21

answered Feb 05 '14 at 19:58

DonAntonio

211,718
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Why can we simply ignore the $a_{k}$ within the summation? – user2850514 Feb 05 '14 at 20:02
Because I thought it was $;f(x)=1;$ for $;x\in\Bbb Q;$, @user2850514. But it's possible to mend this. I shall do this later as I've no time now. – DonAntonio Feb 06 '14 at 00:27
Have you been able to fix this? – user2850514 Feb 06 '14 at 09:38
@user2850514, I have now. Check it. – DonAntonio Feb 06 '14 at 10:21

Working out Riemann sums

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